Based on the information provided, we know that the original solid sample weighed 101.96 g and was composed of 53.96 g of aluminum and the remaining mass as oxygen.
First, let's calculate the amount of oxygen in the initial sample:
\[ \text{Mass of Oxygen} = \text{Total mass of solid} - \text{Mass of Aluminum} \] \[ \text{Mass of Oxygen} = 101.96 , \text{g} - 53.96 , \text{g} = 48.00 , \text{g} \]
Now we know the composition of the solid: 53.96 g of aluminum and 48.00 g of oxygen.
Next, we want to determine the ratio of aluminum to oxygen in the original solid:
\[ \text{Ratio of Aluminum to Oxygen} = \frac{\text{Mass of Aluminum}}{\text{Mass of Oxygen}} = \frac{53.96 , \text{g}}{48.00 , \text{g}} \approx 1.12416 \]
To scale up the solid to 1,937 grams, we need to maintain the same ratio of aluminum to oxygen. The total mass of the new solid required is 1,937 grams.
We'll denote the mass of aluminum as \(x\) and the mass of oxygen as \(y\). The following equations define the relationships:
- \(x + y = 1,937\)
- The ratio should remain approximately \( \frac{53.96}{48.00} \), which can be expressed as \( \frac{x}{y} = \frac{53.96}{48.00} \).
To express \(y\) in terms of \(x\) using the ratio:
\[ y = \frac{48.00}{53.96} x \]
We can then substitute this expression for \(y\) into the first equation:
\[ x + \frac{48.00}{53.96} x = 1,937 \]
Combining both \(x\) terms gives:
\[ x \left(1 + \frac{48.00}{53.96}\right) = 1,937 \]
Calculating \(1 + \frac{48.00}{53.96}\):
\[ 1 + \frac{48.00}{53.96} = 1 + 0.8896 \approx 1.8896 \]
Now we can express the equation as:
\[ x (1.8896) = 1,937 \]
Solving for \(x\):
\[ x = \frac{1,937}{1.8896} \approx 1,025.188 , \text{g (of Aluminum)} \]
Now, substituting \(x\) back to find \(y\):
\[ y = 1,937 - x = 1,937 - 1,025.188 \approx 911.812 , \text{g (of Oxygen)} \]
Thus, in order to create 1,937 grams of the solid, the chemist should react approximately 1,025.19 g of aluminum with approximately 911.81 g of oxygen.
Final Answers:
- Aluminum: 1,025.19 g
- Oxygen: 911.81 g
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